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\begin{document}

\title{MTH 1420, Spring 2012\\Dr. Graham-Squire}
\author{Section 5.5: Integration by Substitution }
\date{}
\maketitle

HW: 6, 12, 18, 26, 31, 34, 42, 47, 63

Practice: 5, 9, 13, 19, 21, 29, 39, 45, 53, 55 

\section{Introduction}

So far we have dealt with antiderivatives only for rather straightforward functions.  Finding antiderivatives becomes much more complicated when we are dealing with functions where you must apply the Product or Chain rules.  In this section, we look at a method for doing the Chain rule backwards, known as the method of substitution.

\bexa
	Let $f(x)=(6x^2+1)^5$.
\eexa  
\vspace{1.5in}

\bexe
	Try to ``think backwards" in the same way as above to find the indefinite integrals below.  You can check your answer by taking the derivative.
	\bite
	\item $\d \int 6x\sin(3x^2-7)\,dx$ 
	\vspace{1in}
	\item $\d \int \frac{(\ln x)^3}{x} \,dx$
	\eite
\eexe

\newpage


\section{The Method of Substitution}

The big idea is this: $\frac{d}{dx} f(g(x))=\blank{2}$ by the chain rule.

This implies that: 

Our goal is to write our integrand in this form somehow.  The tricky part is often to find what the ``inner" function is.

\[\begin{array}{l|l}
\undr{\text{Method of Substitution}} & \text{\undr{Example}}: \d\int x\sin (3x^2-7)\,dx\\
1) \text{ Let } u=g(x), \text{ where } g(x) \text{ is the ``inner function"}.& 1)\\
 & \\
2) \text{ Find } du=g'(x)dx, \text{ and solve for }dx. & 2) \\
 & \\
3) \text { Substitute } u \text{ and } du \text{ back into the integral,} & 3)\\
\text{canceling until you have only } u\text{'s left.} & \\
 & \\
  & \\
4) \text{ Evaluate the integral.} & 4) \\
 & \\
5) \text{ Replace } u \text{ by } g(x). & 5)\\  
\end{array}
\]

\vspace{.2in}
\bexa
	Suppose $\d f'(u)= \frac{(\ln u)^3}{u}$ and $f(1)=2$.  Find $f$.
\eexa
\vspace{2.5in}

\newpage


\bexe
	Find $\d \int e^{7x}dx$.
\eexe
\vspace{2.5in}

Sometimes it is not initially clear how you can use substitution to take an integral, and you have to mess around with the expression a bit.

\bexa
	Calculate $\d\int \tan x\, dx$.
\eexa
\vspace{2.5in}

\section{Substitution with Definite Integrals}

The method for definite integrals is similar, but we have to be careful with our limits of integration.  I will teach you two ways to do it, the bad way and the good way.

\tb{The Bad Way}:  For this method, you act as if you are taking an indefinite integral.  Use substitution just like we did before to get an antiderivative (making sure you substitute back in for $u$!!!!!!) and then evaluate the integral using the limits of integration.
\newpage


\bexa
	$\d\int^1_0\frac{e^z+1}{e^z+z}\, dz$
\eexa
\vspace{3in}

\tb{The Good Way}:  Here you leave the limits of integration on the integral, but when you substitute in the $u$, you \undr{change the limits of integration}to be for $u$, not $x$!  Then solve that definite integral as normal.

\bexa
	$\d\int^1_0\frac{e^z+1}{e^z+z}\, dz$
\eexa
\vspace{3in}
\newpage


\bexe
	Find $\d\int_0^7\sqrt{4+3x} \, dx$.
\eexe
\vspace{3in}

Sometimes the substitution is not even remotely straightforward.

\bexa
	Find $\d\int^1_{-1} v^2(1-v)^6  \,dx$.
\eexa
\vspace{3in}
\newpage


\bexe
	Find $\d\int^1_0 30x(e^{3x^2+7})^5  \,dx$.
\eexe
\vspace{4in}

\bexe
	Find $\d\int^{\pi/6}_{-\pi/6} \tan^3\theta   \,d\theta$.
\eexe
\vspace{3in}

%\theendnotes
\end{document} 